Mathematics - Quadratic Equation Question with Solution | TestHub

Given, $\alpha$ and $\beta$ are the roots of $x^2+2px+q=0$. $$\begin{array}{rlrl}& \therefore & \alpha +\beta & =-2p\\ \text{ and }& \alpha \beta & =q& \end{array}$$ $\alpha$ and $\frac{1}{\beta }$ are the roots of $a{x}^2+2bx+c=0$ and $$\alpha +\frac{1}{\beta }=\frac{-2b}{a}$$ and $$\frac{\alpha }{\beta }=\frac{c}{a}$$ $$\text{ Now, }\begin{array}{rl}\left(p^2-q\right)\left(b^2-ac\right)& =\left[{\left(\frac{\alpha +\beta }{-2}\right)}^2-\alpha \beta \right]\left[{\left(\frac{\alpha +\frac{1}{\beta }}{2}\right)}^2-\frac{\alpha }{\beta }\right]a^2\\ & =\frac{(\alpha -\beta {)}^2}{16}{\left(\alpha -\frac{1}{\beta }\right)}^2\cdot a^2\ge 0\end{array}$$ Statement 1 is true. $$\begin{array}{ll}\text{ Again now, }& pa=-\left(\frac{\alpha +\beta }{2}\right)a=-\frac{a}{2}(\alpha +\beta )\\ \text{ and }& b=-\frac{a}{2}\left(\alpha +\frac{1}{\beta }\right)\\ & pa\ne b\Rightarrow \alpha +\frac{1}{\beta }\ne \alpha +\beta \\ \Rightarrow & \beta \ne 1\\ \because & {\beta }^2\ne \{-1,0,1\},\text{ correct. }\\ \text{ Similarly, if }& c\ne qx\\ \Rightarrow & a\frac{\alpha }{\beta }\ne a\alpha \beta \Rightarrow \alpha \left(\beta -\frac{1}{\beta }\right)\ne 0\end{array}$$ $$\begin{array}{ll}\Rightarrow & \alpha \ne 0\text{ and }\beta -\frac{1}{\beta }\ne 0\\ \Rightarrow & \beta \ne \{-1,0,1\}\end{array}$$ Statement 2 is true. Both Statement 1 and Statement 2 are true. But Statement 2 does not explain Statement 1.